1. Standard Enthalpy of Formation Chapter 5.5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Energy & Rates of Reaction Unit 3

2. 5.5: Standard enthalpy of formation f Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.  H 0 (O 2 ) = 0  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f 5.5

4. The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - 5.5 Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

5. Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g)  H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H 0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 5.5

6. Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - H0H0 rxn 6  H 0 (H 2 O) f 12  H 0 (CO 2 ) f = [+] - 2  H 0 (C 6 H 6 ) f [] H0H0 rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ - 6534.9 kJ 2 mol = - 3267.44 kJ/mol C 6 H 6 5.5

7. The enthalpy of solution (  H soln ) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.  H soln = H soln - H components 6.6 Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? LiCl & CaCl 2 NaCl, KCl, NH 4 Cl, NH 4 NO 3

8. Sample Problem Calculate the heat of combustion of methanol. CH 3 OH (l) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O (g)  H 0 comb = ?

9. Determine ∆H˚ r for the following reaction using the enthalpies of formation that are provided. UNIT 3 Section 5.3 TO PREVIOUS SLIDEPREVIOUS Chapter 5: Energy Changes C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(l) ∆H˚ f of C 2 H 5 OH(l): –277.6 kJ/mol ∆H˚ f of CO 2 (g): –393.5 kJ/mol ∆H˚ f of H 2 O(l): –285.8 kJ/mol L EARNING C HECK

10. Multistep Energy Calculations Several energy calculations may be required to solve multistep problems –Heat flow q = mcΔT –Enthalpy changes ΔH = n ΔH x –Hess’s Law ΔH target = ∑ΔH known –Enthalpies of Formation

11. Example Ans: 0.13kg